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For example, here is an excerpt from "Introduction to manifolds" by loring W. Tu

20.4. If X and Y are smooth vector fields on a manifold M, then the Lie derivative $\ \mathcal{L}_XY$ coincides with the Lie bracket [X,Y].

Proof. It suffices to check the equality $\ \mathcal{L}_XY$ = [X,Y] at every point

Why do we need to check it at every point? Other proofs suffices for some arbitrary points:

Proof. In this proposition, there are three operations—exterior differentiation, differentiation with respect to t, and evaluation at $\ t = t_0$. We will first show that d and d/dt commute:

$\ \frac d{dt}(dω_t) = d \frac d{dt} ω_t $

It is enough to check the equality at an arbitrary point p ∈ M.

Isn't checking at an arbitrary point the same as checking at every point?

folo polo
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    One checks at an arbitrary point. That shows it holds at every point. Please give an example of what you mean by "suffices for some points." – Ted Shifrin Nov 20 '19 at 19:28

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They are equivalent statements. It APPEARS that we do less work in one version, but we don't. Making sure that a point picked arbitrarily (for you by someone else) satisfies a certain property is the same as showing that all points satisfy that property.

Claiming that any one ball picked randomly from a bag is green is the same claim that ALL balls in the bag are green. We cannot just afford an exception.

Behnam Esmayli
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  • Welcome. So, please "accept" this as an answer if this solves your question! (By clicking on the checkmark to the left of the paragraphs of the question.) – Behnam Esmayli Nov 22 '19 at 19:01